Embed Longest Rings onto Star Graphs with Vertex Faults
Sun-Yuan Hsieh, Gen-Huey Chen
Department of Computer Science and Information Engineering
National Taiwan University, Taipei, Taiwan, ROC
and
Chin-Wen Ho
Department of Computer Science and Information Engineering
National Central University, Taipei, Taiwan, ROC
Abstract
The star graph has been recognized as an attractive alternative to the hypercube. Let Fe and Fv be the sets of vertex faults and edge faults, respectively. Previously, Tseng et al. showed that an n-dimensional star graph can embed a ring of length n! if |Fe |≤n-3 (|Fv|=0), and a ring of length at least n!-4|Fv | if |Fv |≤n-3 (|Fe |=0). Since an n-dimensional star graph is regular of degree n-1 and is bipartite with two partite sets of equal size, our result achieves optimality in the worst case.
1 Introduction
One crucial step on designing a large-scale multiprocessor system is to determine the topology of the interconnection network (network for short). In recent decades, a lot of networks have been proposed. The interested readers may refer to [4, 17] for extensive references. Among them, the star graph [1], which belongs to the class of Cayley graphs [2], has been recognized as an attractive alternative to the hypercube. It possesses many nice topological properties such as recursiveness, symmetry, maximal fault tolerance, sublogarithmic degree and diameter, and strong resilience, which are all desirable when we are building interconnection topologies for parallel and distributed systems. Besides, the star graph can embed rings [25], meshes [27], trees [3], and hypercubes [23]. Efficient communication algorithms for shortest-path routing [1], multiple-path routing [12], multicasting [16], broadcasting [31], gossiping [5], and scattering [16] have been proposed. Efficient algorithms for sorting [22], merging [25], selection [24], prefix sums [25], ranking [30], Fourier
transform [15], and computational geometry [26] have also been designed.
Since processor and/or link faults may happen when a network is put in use, it is practically meaningful to consider faulty networks. The problems of diameter [28], routing [14, 34], multicasting [21], broadcasting [33], gossiping [13], embedding [11, 29, 35], and fault-tolerant graphs [6-9, 19] have been solved on a variety of faulty networks. This paper is concerned with the problem of embedding a ring onto a faulty star graph. Throughout this paper, we use network and graph, processor and vertex, and link and edge, interchangeably.
Previously, embedding rings onto faulty star graphs has been studied in [20, 32]. In [20], Latifi and Bagherzadeh showed that an n-dimensional star graph with vertex faults can embed a ring of length n!-m!, where all vertex faults belong to an m-dimensional star graph and
m≤n is minimal. Letting Fv and Fe denote the sets of vertex
faults and edge faults, respectively, Tseng et al. [32] showed that an n-dimensional star graph can embed a ring of length n! if |Fe|≤n-3 (|Fv|=0), and a ring of length at least n!-4|Fv| if |Fv|≤n-3 (|Fe|=0). Since the n-dimensional star
graph is regular of n-1 and contains n! vertices, the former is optimal in the worst case. However, the latter is not optimal.
In this paper, we improve Tseng et al.’s latter result by lengthening the embedded ring from n!-4|Fv| to n!-2|Fv|.
We note that the star graph is bipartite and has two partite sets of equal size [18]. When |Fv| vertex faults fall into the
same partite set, an embedded ring has length not more than n!-2|Fv|. Hence, our result achieves optimality in the
2 Preliminaries
In this section the structure of the star graph is formally described. Some necessary notations and basic operations are also introduced. We use Sn to denote an n-dimensional
star graph.
Definition 1. The vertex set of Sn is denoted by
{a1a2...an | a1a2...an is a permutation of 1, 2, ..., n}. Vertex
adjacency is defined as follows: a1a2...an is adjacent to aia2...ai-1a1ai+1...an for all 2≤i≤n.
The vertices of Sn are n! permutations of 1, 2, ..., n,
and there is an edge between two vertices of Sn if and only
if they can be obtained from each other by swapping the leftmost number with one of the other n-1 numbers. We refer to the position of ai in a1a2...an as the ith dimension,
and the edge between a1a2...an and aia2...ai-1a1ai+1...an as
the ith-dimensional edge. Clearly Sn is regular of degree
n-1. Since S1 is a vertex, S2 is an edge, and S3 forms a cycle
of length 6, we consider Sn with n≥4 throughout this paper.
There are embedded Sr’s contained in Sn, where 1≤r≤n.
An embedded Sr can be conveniently represented by
<s1s2...sn>r, where s1=∗, si ∈{∗, 1, 2, ..., n} for all 2≤i≤n,
and exactly r of s1, s2, ..., sn are ∗ (∗ denotes a “don’t care”
symbol). For example, <∗∗∗3>3, which represents an
embedded S3 in S4, contains six nodes 1243, 1423, 2143,
2413, 4123, and 4213. In terms of graph, <∗∗∗3>3 is a
subgraph of S4 induced by {1243, 1423, 2143, 2413, 4123,
4213}. When r=n, <s1s2...sn>n represents Sn. Two basic
operations on Sn are defined as follows.
Definition 2. An i-partition on <s1s2...sn>r partitions <s1s2...sn>r into r embedded Sr-1’s, denoted by <s1s2...s i-1qsi+1...sn>r-1, where 2≤i≤n, si=∗, and q ∈{1, 2, ..., n}-{s1, s2, …, sn}.
For example, executing a 3-partition on <∗∗∗15>3
produces three embedded S2’s, i.e., <∗∗215>2, <∗∗315>2,
and <∗∗415>2.
Definition 3. An (i1, i2, ..., im)-partition on <s1s2...sn>r
performs an i1-partition, an i2-partition, ..., an im-partition,
sequentially, on <s1s2...sn>r, where i1i2...im is a permutation
of m elements from {2, 3, ..., n} and
s
i1=
s
i2= ⋅⋅⋅ =s
im=∗.After executing an (i1, i2, ..., im)-partition, <s1s2...sn>r is
partitioned into r(r-1)⋅⋅⋅(r-m+1) embedded Sr-m’s.
For example, when a (3, 2)-partition is applied to <∗∗∗15>3, a 3-partition is first executed on <∗∗∗15>3 to
produce three embedded S2’s, i.e., <∗∗215>2, <∗∗315>2,
and <∗∗415>2. Then, a 2-partition is executed on each of
these three S2’s to produce six embedded S1’s, i.e.,
<∗3215>1, <∗4215>1, <∗2315>1, <∗4315>1, <∗2415>1,
and <∗3415>1.
Two embedded Sr’s <s1s2...sn>r and <t1t2...tn>r are said
to be adjacent if sj≠∗, tj≠∗, and sj≠tj for some 2≤j≤n, and si=ti for all 1≤i≤n and i≠j. Moreover, the position j is
denoted by dif(<s1s2...sn>r, <t1t2...tn>r). For example,
<∗∗23>2 is adjacent to <∗∗13>2, and dif(<∗∗23>2,
<∗∗13>2)=3.
In the rest of this paper, an embedded Sr is named an r-vertex when it is regarded as a supervertex. Imaginarily,
there is a superedge, named an r-edge, between two adjacent r-vertices. Really, an r-edge in Sn comprises (r-1)!
edges. A 1-vertex in Sn is a vertex of Sn, and a 1-edge in Sn
is an edge of Sn.
Definition 4. Suppose A0, A1, ..., An(n-1)(n-2)⋅⋅⋅(r+1)-1 are n(n-1)(n-2)⋅⋅⋅(r+1) r-vertices that result from executing an (i1, i2, ..., in-r)-partition on Sn, where 1≤r≤n-1. They form an r-ring, denoted by Rr=[A0, A1, ..., An(n-1)(n-2)⋅⋅⋅(r+1)-1], if Ai is
adjacent to A(i-1) mod n(n-1)(n-2)⋅⋅⋅(r+1) and A(i+1) mod n(n-1)(n-2)⋅⋅⋅(r+1)
for all 0≤i≤n(n-1)(n-2)⋅⋅⋅(r+1)-1.
Definition 5. An i-partition on an Rr=[A0, A1, ..., A
n(n-1)(n-2)⋅⋅⋅(r+1)-1] performs an i-partition on A0, A1, ..., An1)
(n-2)⋅⋅⋅(r+1)-1, where 2≤i≤n and r≥2. An i-partition on an Rr is
abbreviated to a partition on an Rr if the position i is
“don’t care”.
After executing an i-partition on an Rr=[A0, A1, ..., A n(n-1)(n-2)⋅⋅⋅(r+1)-1], each Aj is partitioned into r (r-1)-vertices,
where 0≤j≤n(n-1)(n-2)⋅⋅⋅(r+1)-1. Since every two of the r (r-1)-vertices are adjacent, each Aj can be viewed as a
complete graph of r (r-1)-vertices. Throughout this paper, we use
K
rr−1 to denote the complete graph.Suppose Ak=<s1...si-1sisi+1...sj-1xsj+1...sn>r and A(k+1) mod
n(n-1)(n-2)⋅⋅⋅(r+1)=<s1...si-1sisi+1...sj-1ysj+1...sn>r are two
neighboring r-vertices in an Rr, where r≥2, x≠y, and si=∗.
After an i-partition, they each are partitioned into r (r-1)-vertices. Among them, r-1 from Ak are adjacent to r-1 from A(k+1) mod n(n-1)(n-2)⋅⋅⋅(r+1). The other two that are not adjacent
are <s1...si-1ysi+1...sj-1xsj+1...sn>r-1 and <s1...si-1xsi+1...sj-1ysj+1... sn>r-1.
Suppose an Rk=[A0, A1, ..., An(n-1)(n-2)⋅⋅⋅(k+1)-1] is obtained
in Sn, where k≥4. After executing a partition on the Rk, each Ai forms a
K
kk−1
, where 0≤i≤n(n-1)⋅⋅⋅(k+1)-1. There exist two (k-1)-vertices, say Xi and Yi, in each Ai so that Xi and Yi
are adjacent to Y(i-1) mod n(n-1)⋅⋅⋅(k+1) and X(i+1) mod n(n-1)⋅⋅⋅(k+1),
respectively. We refer to Xi and Yi as the entry (k-1)-vertex and exit (k-1)-vertex of Ai, respectively, in subsequent
3 Embed a longest ring onto S
nwith vertex
faults
In this section we assume that Sn has |Fv|≤n-3 vertex faults.
We say that an i-vertex is healthy if it has no vertex fault, and faulty otherwise, where 1≤i≤n. Similarly, a ring (or a
path) is healthy if it has no vertex fault. In this section we show that Sn with n≥4 can embed a healthy ring of length n!-2|Fv|. Since Sn is bipartite with two partite sets of equal
size, the length n!-2|Fv| is maximum in the worst case.
Moreover, |Fv|=n-3 is also maximum in the worst case in
order to embed a healthy ring of maximum length, because
Sn is regular of n-1. Our method is first to determine an R4
in Sn that owns the following three properties:
(P1) each 4-vertex of the R4 has at most one vertex
fault;
(P2) for every three consecutive 4-vertices
U=<u1u2…un>4, V=<v1v2…vn>4, and W=<w1w2…wn>4 in the R4, udif(U,V)≠wdif(V,W)
holds;
(P3) both every two consecutive 4-vertices in the
R4 are not faulty.
Then, a healthy ring of length n!-2|Fv| can be generated
from the R4.
Lemma 1. Suppose U=<u1u2…un>r, V=<v1v2…vn>r,
and W=<w1w2…wn>r are arbitrary three consecutive
r-vertices in an Rr, where r≥2. Let p=dif(U, V) and q=dif(V, W). If up≠wq, then after executing a partition on the Rr,
each (r-1)-vertex of V is connected to U or W.
Proof. Without loss of generality, we assume that a
j-partition is executed on the Rr, where 2≤j≤n. Hence, uj=vj=wj=∗. Since p=dif(U, V)≠1 and q=dif(V, W)≠1, we
have up≠vp, vq≠wq, ui=vi for all 1≤i≤n and i≠p, and vi=wi for
all 1≤i≤n and i≠q. Suppose conversely that up≠wq and there
exists an (r-1)-vertex, say V1=<v1v2…vj-1zvj+1…vn>r-1, of V,
which is connected to neither of U and W. Thus, z=up, for
otherwise V1 is adjacent to some (r-1)-vertex of U.
Similarly, z=wq. Hence, up=z=wq, which contradicts our
assumption. Q.E.D.
The following lemma was shown in [32].
Lemma 2.[32] Suppose |Fv|≤n-3. There exists a
sequence a1, a2, …, an-4 of positions so that after executing
an (a1, a2, …, an-4)-partition on Sn, each resulting 4-vertex
contains at most one vertex fault.
The positions a1, a2, …, an-4 in Lemma 2 can be easily
determined as follows. We let a1 be a position where at
least two vertices in Fv differ. For example, if Fv={123456,
123654}, a1 is set to 4 or 6. An a1-partition is then
executed on Sn to produce n (n-1)-vertices, and Fv is
partitioned accordingly. That is, two vertices in Fv fall into
the same subset if and only if they belong to the same (n-1)-vertex. The position a2 is determined similarly. We
simply let a2 be a position where at least two vertices in
some subset differ. Then an a2-partition is executed on the n (n-1)-vertices to produce n(n-1) (n-2)-vertices, and every
non-empty subset of Fv is partitioned accordingly. The
process is repeated until every non-empty subset contains one vertex, when the remaining positions are determined arbitrarily.
A path in a graph G is said to be a hamiltonian path if it contains every vertex of G exactly once [10]. In the following discussion, a path between two vertices, say X and Y, are abbreviated to an X-Y path.
Lemma 3. Suppose n≥6 and |Fv|≤n-3. An R4 with
properties (P1), (P2), and (P3) can be generated in Sn.
Proof. Suppose a1, a2, …, an-4 are a sequence of
positions that satisfy Lemma 2. We first construct an R6 in Sn. When n=6, Sn is an R6 with only one 6-vertex. For n>6,
an R6 can be obtained by executing an (a1, a2, …, an-6
)-partition on Sn as follows. Initially, an a1-partition is
applied to Sn, and so a
K
n n−1results. An Rn-1 can be easily
generated from the
K
nn−1. Then, for j=2, 3, …, n-6, an R n-j can be generated from an Rn-j+1 as explained below.Suppose the Rn-j+1=[An-j+1,0, An-j+1,1, …, An-j+1,n(n-1)⋅⋅⋅ (n-j+2)-1]. Each An-j+1,k forms a
K
n jn j
− +− 1 after an aj-partition is
executed on the Rn-j+1, where 0≤k≤n(n-1)⋅⋅⋅(n-j+2)-1. Let Xk
and Yk denote a pair of distinct entry and exit (n-j)-vertices
of An-j+1,k. It is easy to establish a hamiltonian Xk-Yk path in
each
K
n jn j− +− 1 formed by An-j+1,k. Then an Rn-j can begenerated if all the hamiltonian paths are interleaved with (n-j)-edges (Y0, X1), (Y1, X2), …, (Yn(n-1)⋅⋅⋅(n-j+2)-1, X0). When j=n-6, an R6=[A6,0, A6,1, …, A6,n(n-1)⋅⋅⋅7-1] is obtained.
We continue to generate an R5 from the R6. When n=6,
it is easy to generate an R5 so that for arbitrary three
consecutive 5-vertices U=<u1u2…un>5, V=<v1v2…vn>5, and W=<w1w2… wn>5 in the R5, udif(U,V)≠wdif(V,W) holds. When n>6, each A6,r forms a
K
65
after an an-5-partition is
executed on the R6, where 0≤r≤n(n-1)⋅⋅⋅7-1. Let Xr and Yr
denote a pair of distinct entry and exit 5-vertices of A6,r. It
is easy to establish a hamiltonian Xr-Yr path in the
K
6 5formed by A6,r whose first (last) two 5-vertices are
connected to A6,(r-1) mod n(n-1)⋅⋅⋅7 (A6,(r+1) mod n(n-1)⋅⋅⋅7). All the
X2), …, (Yn(n-1)⋅⋅⋅7-1, X0) form an R5. Besides, for arbitrary
three consecutive 5-vertices U=<u1u2…un>5, V=<v1v2…vn>5, and W=<w1w2… wn>5 in the R5, udif(U,V)≠
wdif(V,W) holds. The reason is explained below.
Let p=dif(U, V) and q=dif(V, W). If U=Yi for some
0≤i≤n(n-1)(n-2)⋅⋅⋅7-1, then V=X(i+1) mod n(n-1)⋅⋅⋅7 and W is the
second 5-vertex in the hamiltonian X(i+1) mod n(n-1)⋅⋅⋅7-Y(i+1) mod
n(n-1)⋅⋅⋅7 path, which cause p≠an-5 and q=an-5. Moreover, W is
connected to A6,i according to the construction above.
Recall that the 5-vertex of A6,i that are not connected to A6,(i+1) mod n(n-1)⋅⋅⋅7 is <x1...xq-1ypxq+1...xn>5 and the 5-vertex of A6,(i+1) mod n(n-1)⋅⋅⋅7 that are not connected to A6,i is <y1...y q-1xpyq+1...yn>5, where xq=yp≠xp=yq and xi=yi for all 1≤i≤n and i∉{p, q}. Hence, if wq=up, then W is not connected to A6,i,
which is a contradiction. If W=Xi for some 0≤i≤
n(n-1)(n-2)⋅⋅⋅7-1, then up≠wq can be shown similarly. Otherwise, if
U, V, and W belong to the same 6-vertex, then up≠wq
because p=q=an-5.
We note that at most one 5-vertex in the R5 contains
two vertex faults and the others each contain at most one vertex fault. If there are two 5-vertices in the R5 that each
contain two or more vertex faults, then at most n-5 of the 5-vertices in the R5 are faulty. This is a contradiction
because the R5, which results from executing an (a1, a2, ..., an-5)-partition on Sn, contains at least n-4 faulty 5-vertices
(see the paragraph after Lemma 2).
Next, we generate a desired R4 from the R5. Suppose
the R5=[A5,0, A5,1, …, A5,n(n-1)⋅⋅⋅6-1]. Without loss of
generality, we assume that one 5-vertex, say A5,k (0≤k≤
n(n-1)(n-2)⋅⋅⋅6-1), of the R5 contains two vertex faults and the
others each contain at most one vertex fault. An an-4
-partition is executed on the R5, and so each A5,j forms a
K
54, where 0≤j≤n(n-1)⋅⋅⋅6-1. According to Lemma 1, each 4-vertex of A5,k is connected to A5,(k-1) mod n(n-1)⋅⋅⋅6 or A5,(k+1) mod n(n-1)⋅⋅⋅6. We further assume, without loss of generality,that A5,k contains two faulty 4-vertices, say C and E. A
hamiltonian path in the
K
54 formed by A5,k can beestablished according to the following three cases.
Case 1. Both of C and E are connected to both A5,(k-1) mod
n(n-1)⋅⋅⋅6 and A5,(k+1) mod n(n-1)⋅⋅⋅6. Let B (F) be the 4-vertex in
A5,k that is not connected to A5,(k+1) mod n(n-1)⋅⋅⋅6 (A5,(k-1) mod n(n-1)⋅⋅⋅6). A hamiltonian path is established as (B, C, D, E, F),
where D is the other 4-vertex of A5,k.
Case 2. One of C and E is connected to both A5,(k-1) mod n(n-1)⋅⋅⋅6 and A5,(k+1) mod n(n-1)⋅⋅⋅6, and the other is connected to one
of A5,(k-1) mod n(n-1)⋅⋅⋅6 and A5,(k+1) mod n(n-1)⋅⋅⋅6. Without loss of
generality, we assume C is connected to both A5,(k-1) mod n(n-1)⋅⋅⋅6 and A5,(k+1) mod n(n-1)⋅⋅⋅6. If E is connected to A5,(k+1) mod n(n-1)⋅⋅⋅6, a hamiltonian path is established as (B, C, D, E, F),
where B is the 4-vertex of A5,k that is not connected to A5, (k+1) mod n(n-1)⋅⋅⋅6, and D and F are the other two 4-vertices of A5,k. If E is connected to A5,(k-1) mod n(n-1)⋅⋅⋅6, a hamiltonian
path is established as (D, E, F, C, B), where B is the 4-vertex of A5,k that is not connected to A5,(k-1) mod n(n-1)⋅⋅⋅6, and D and F are the other two 4-vertices of A5,k.
Case 3. One of C and E is not connected to A5,(k-1) mod n(n-1)⋅⋅⋅6,
and the other is not connected to A5,(k+1) mod n(n-1)⋅⋅⋅6 (it is
impossible that C and E are not connected to A5,(k-1) mod n(n-1)⋅⋅⋅6 or A5,(k+1) mod n(n-1)⋅⋅⋅6). Without loss of generality, we
assume C is not connected to A5,(k+1) mod n(n-1)⋅⋅⋅6 and E is not
connected to A5,(k-1) mod n(n-1)⋅⋅⋅6. A hamiltonian path is
established as (B, C, D, E, F), where B, D, and F are the other three 4-vertices of A5,k.
The first (last) 4-vertex in the hamiltonian paths above is the entry (exit) 4-vertex of A5,k. Besides, the first (last)
two 4-vertices are connected to A5,(k-1) mod n(n-1)⋅⋅⋅6 (A5,(k+1) mod
n(n-1)⋅⋅⋅6). Then we determine the entry and exit 4-vertices of
A5,j, denoted by X5,j and Y5,j, respectively, for all 0≤j≤
n(n-1)(n-2)⋅⋅⋅5-1 and j≠k so that both Y5,(j-1) mod n(n-1)⋅⋅⋅6 and X5,j
are not faulty and both Y5,j and X5,(j+1) mod n(n-1)⋅⋅⋅6 are not
faulty. It is not difficult to establish a hamiltonian X5,j-Y5,j
path in the
K
54 formed by each A5,j whose first (last) two4-vertices are connected to A5,(j-1) mod n(n-1)⋅⋅⋅6 (A5,(j+1) mod n(n-1)⋅⋅⋅6).
All the hamiltonian paths interleaved with 4-edges constitute an R4. Clearly the R4 has properties (P1) and
(P3). Recall that for arbitrary three consecutive 5-vertices
U=<u1u2…un>5, V=<v1v2…vn>5, and W=<w1w2…wn>5 in
the R5, udif(U,V)≠wdif(V,W) holds. With similar arguments, the R4 has property (P2). This completes the proof.
Q.E.D.
Lemma 4. Suppose |Fv|≤1. There is a healthy path of
maximal length 4!-3 between arbitrary two adjacent healthy vertices of S4.
Proof. Suppose u and v are arbitrary two adjacent
healthy vertices of S4, and let f=f1f2f3f4 denote the faulty
vertex. Since the star graph is edge symmetric [2], we assume u=u1u2u3u4=1234 and v=v1v2v3v4=3214 for
illustration. We have f2≠u2=v2 or f4≠u4=v4, for otherwise f=u or f=v, which is a contradiction. We assume f4≠u4=v4=4,
without loss of generality. After executing a 4-partition on
S4, u and v belong to <∗∗∗4>3 and f belongs to<∗∗∗1>3 or
<∗∗∗2>3 or <∗∗∗3>3.
If f ∈<∗∗∗1>3, a healthy u-v path of length 4!-3 can be
constructed as follows:
(u=1234, 2134, 3124, 1324, 4321, 3421, 2431, 4231, 3241, 1243, 2143, 4123, 1423, 2413, 3412, 1432,
4132, 3142, 1342, 4312, 2314, 3214=v) if f=2341; (u=1234, 4231, 3241, 2341, 4321, 3421, 1423, 2413, 4213, 1243, 2143, 3142, 1342, 4312, 3412, 1432, 4132, 2134, 3124, 1324, 2314, 3214=v) if f=2431; (u=1234, 4231, 2431, 3421, 4321, 2341, 1342, 4312, 3412, 1432, 4132, 3142, 2143, 1243, 4213, 2413, 1423, 4123, 3124, 1324, 2314, 3214=v) if f=3241; (u=1234, 2134, 3124, 4123, 1423, 2413, 4213, 1243, 2143, 3142, 1342, 4312, 3412, 1432, 2431, 4231, 3241, 2341, 4321, 1324, 2314, 3214=v) if f=3421; (u=1234, 2134, 4132, 3142, 1342, 4312, 3412, 1432, 2431, 3421, 4321, 2431, 3241, 1243, 4213, 2413, 1423, 4123, 3124, 1324, 2314, 3214=v) if f=4231; (u=1234, 2134, 3124, 4123, 2143, 1243, 4213, 2413, 1423, 3421, 2431, 4231, 3241, 2341, 1342, 3142, 4132, 1432, 3412, 4312, 2314, 3214=v) if f=2341.
If f ∈<∗∗∗2>3 or <∗∗∗3>3, a healthy u-v path of length
4!-3 can be obtained similarly. Since the star graph is bipartite, the u-v path is the longest in the worst case. Q.E.D.
We note that S3 forms a ring of length six. The
following two lemmas were shown in [32].
Lemma 5.[32] Suppose U and V are two adjacent 3-vertices in Sn, and let (c0, c1, …, c5) denote the ring formed
by U. Then, the vertices of U that are connected to V are cj
and c(j+3) mod 6 for some 0≤j≤5.
Lemma 6.[32] Suppose U=<u1u2…un>3, V=<v1v2…vn>3, and W=<w1w2…wn>3 are three 3-vertices
in Sn, and V is adjacent to both U and W. If udif(U,V)≠wdif(V,W),
then the two vertices of V that are connected to U are disjoint from the two vertices of V that are connected to W.
Lemma 7. Suppose n≥5 and |Fv|≤n-3. A healthy ring of
maximal length n!-2|Fv| can be generated from an R4 in Sn
that owns properties (P1), (P2), and (P3).
Proof. Suppose an R4=[A0, A1, …, An(n-1)⋅⋅⋅5-1] with
properties (P1), (P2), and (P3) is obtained after executing an (a1, a2, …, an-4)-partition on Sn. An an-3-partition is then
executed on the R4, and so each Ai is partitioned into four
3-vertices that form a
K
43, where an-3 ∈{2, 3, …, n}-{a1, a2, …, an-4} and 0≤i≤n(n-1)⋅⋅⋅5-1. According to Lemma 1,each 3-vertex of Ai is connected to A(i-1) mod n(n-1) ⋅⋅⋅5 or A(i+1) mod n(n-1)⋅⋅⋅5 because of property (P2). Since three 3-vertices
of Ai are connected to A(i-1) mod n(n-1)⋅⋅⋅5 and three 3-vertices
of Ai are connected to A(i+1) mod n(n-1)⋅⋅⋅5, two 3-vertices of Ai
are connected to both A(i-1) mod n(n-1)⋅⋅⋅5 and A(i+1) mod n(n-1)⋅⋅⋅5.
We then determine the entry and exit 3-vertices of each Ai, denoted by Xi and Yi, respectively, as follows. For
each faulty Ai, we let Xi=Yi=Q, where Q is a healthy
3-vertex of Ai that is connected to both A(i-1) mod n(n-1)⋅⋅⋅5 and A(i+1) mod n(n-1)⋅⋅⋅5. We note that Ai contains one vertex fault
(property P(1)). Since A(i-1) mod n(n-1)⋅⋅⋅5 and A(i+1) mod n(n-1)⋅⋅⋅5
are healthy (property (P3)), Y(i-1) mod n(n-1)⋅⋅⋅5 and X(i+1) mod n(n-1)⋅⋅⋅5 are determined as the 3-vertex of A(i-1) mod n(n-1)⋅⋅⋅5 and
the 3-vertex of A(i+1) mod n(n-1)⋅⋅⋅5, respectively, that are
adjacent to Q. For every two consecutive healthy Ai and A(i+1) mod n(n-1)⋅⋅⋅5, Yi and X(i+1) mod n(n-1)⋅⋅⋅5 are determined as
two adjacent 3-vertices that belong to Ai and A(i+1) mod n(n-1)⋅⋅⋅5, respectively.
All Xi and Yi thus determined are healthy. Moreover,
when Xi=Yi,
u
dif Y( (i−1) mod (n n−1)5,Xi)≠w
dif X Xi i n n
( , (+1) mod ( −1)5) holds, where Y(i-1) mod n(n-1)⋅⋅⋅5=<u1u2…un>4 and X(i+1) mod n(n-1)⋅⋅⋅5=<w1w2…wn>4 are
assumed. The reason is explained as follows. Suppose
conversely
u
dif Y Xi n n i
( (−1) mod ( −1)5, )=
w
dif X Xi i n n
( , (+1) mod ( −1)5). If dif(Y(i-1) mod n(n-1)⋅⋅⋅5, Xi)=dif(Xi, X(i+1) mod n(n-1)⋅⋅⋅5), then Y(i-1) mod n(n-1)⋅⋅⋅5=X(i+1) mod n (n-1)⋅⋅⋅5, which is a contradiction. If dif(Y(i-1) mod n(n-1)⋅⋅⋅5, Xi)≠dif(Xi, X(i+1) mod n(n-1)⋅⋅⋅5), then there is a 3-vertex of Ai
which is connected to neither of A(i-1) mod n(n-1)⋅⋅⋅5 and A(i+1) mod n(n-1)⋅⋅⋅5. This is again a contradiction because each
3-vertex of Ai is connected to A(i-1) mod n(n-1)⋅⋅⋅5 or A(i+1) mod n(n-1)⋅⋅⋅5.
Next we generate a healthy ring of length n!-2|Fv| by
establishing healthy paths in A0, A1, …, An(n-1)⋅⋅⋅5-1 in this
sequence. First we construct a healthy path in A0 as follows.
Let x0 be a healthy vertex of X0 which is adjacent to a
healthy vertex of Yn(n-1)⋅⋅⋅5-1. When X0=Y0, ) , (Y( 1) 51X0 dif nn
u
−− ≠
w
dif(X0,X1) holds for three consecutive3-vertices Yn(n-1)⋅⋅⋅5-1=<u1u2…un>4, X0(=Y0), and X1=<w1w2… wn>4. According to Lemmas 5 and 6, there is
a healthy vertex y0 of Y0(=X0) which is adjacent to both x0
and x1, where x1 is a healthy vertex of X1. If A0 is healthy, it
is not difficult to establish a healthy hamiltonian x0-y0 path
in A0. If A0 is faulty, a healthy x0-y0 path of maximal length
4!-3 in A0 can be established according to Lemma 4.
We then consider the situation of X0≠Y0. Let Q0 (L0) be
the 3-vertex of A0 that is not connected to A1 (An(n-1)⋅⋅⋅5-1),
where Q0≠Y0 (L0≠X0). A healthy hamiltonian X0-Y0 path for
Q0=X0 and L0=Y0, (X0, Q0, C, Y0) if Q0≠X0 and L0=Y0, (X0, B, L0, Y0) if Q0=X0 and L0≠Y0, and (X0, Q0, L0, Y0) if Q0≠X0
and L0≠Y0, where B and C are the other 3-vertices of A0.
Without loss of generality, we assume (X0, Q0, L0, Y0)
is the healthy hamiltonian X0-Y0 path. With the same
arguments as the proof of Lemma 3, every three consecutive 3-vertices in {Yn(n-1) ⋅⋅⋅5-1, X0, Q0, L0, Y0, X1}
satisfy property (P2). Since A0 is healthy, according to
Lemmas 5 and 6 there are four pairs of distinct healthy vertices, denoted by {x0, p}, {q, r}, {s, t}, and {z, y0}, of X0, Q0, L0, and Y0, respectively, so that every two consecutive
vertices in {x0, p, q, r, s, t, z, y0, x1} are adjacent, where x1
is a healthy vertex of X1. Clearly, there are healthy
hamiltonian x0-p, q-r, s-t, and z-y0 paths in X0, Q0, L0, and Y0, respectively. These healthy hamiltonian paths
interleaved with edges (p, q), (r, s), and (t, z) constitute a healthy hamiltonian x0-y0 path in A0.
Then we construct healthy paths in A1, A2, …, An(n-1)⋅⋅⋅ 5-2, sequentially. Suppose a healthy xj-1-yj-1 path in Aj-1 is
obtained, where 1≤j≤n(n-1)⋅⋅⋅5-2. Very similar to the situation of A0, a healthy path in Aj can be constructed as
follows. Let xj be the (healthy) vertex of Xj that is adjacent
to yj-1. When Xj=Yj,
u
dif(Yj−1,Xj)≠w
dif(Xj,Xj+1) holds forthree consecutive 3-vertices Yj-1=<u1u2…un>4, Xj(=Yj), and Xj+1=<w1w2…wn>4. According to Lemmas 5 and 6, there is
a healthy vertex yj of Yi(=Xj) which is adjacent to both xj
and a healthy vertex of Xj+1. If Aj is healthy, there is a
healthy hamiltonian xj-yj path in Aj. If Aj is faulty, a healthy xj-yj path of maximal length 4!-3 in Aj can be established
according to Lemma 4. When Xj≠Yj, a hamiltonian Xj-Yj
path for the
K
43 formed by Aj is first established. Then ahealthy hamiltonian xj-yj path in Aj can be obtained by
concatenating healthy hamiltonian paths for the four 3-vertices of Aj, where yj is a healthy vertex of Yj and is
adjacent to a healthy vertex of Xj+1.
Thus far we have established healthy paths in A0, A1, …, An(n-1)⋅⋅⋅5-2, We continue to establish a healthy path in
An(n-1)⋅⋅⋅5-1. Let xn(n-1)⋅⋅⋅5-1 be the (healthy) vertex of Xn(n-1)⋅⋅⋅5-1
that is adjacent to yn(n-1)⋅⋅⋅5-2, and yn(n-1)⋅⋅⋅5-1 be the (healthy)
vertex of Yn(n-1)⋅⋅⋅5-1 that is adjacent to x0. Without loss of
generality, we use (c0, c1, c2, c3, c4, c5) to represent Yn(n-1)⋅⋅⋅ 5-1, where c0=yn(n-1)⋅⋅⋅5-1 is assumed. When Xn(n-1)⋅⋅⋅5-1=Yn(n-1) ⋅⋅⋅ 5-1,
u
dif(Yn(n−1)...5−2,Xn(n−1)...5−1)≠w
dif(Xn(n−1)...5−1,X0) holds forthree consecutive 3-vertices Yn(n-1)⋅⋅⋅5-2=<u1u2… un>4, X
n(n-1)⋅⋅⋅5-1(=Yn(n-1)⋅⋅⋅5-1), and X0=<w1w2…wn>4. According to
Lemmas 5 and 6, xn(n-1)⋅⋅⋅5-1 cannot be c0 or c3. Actually, x n(n-1)⋅⋅⋅5-1=c1 or c5 (i.e., xn(n-1)⋅⋅⋅5-1 and yn(n-1)⋅⋅⋅5-1 are adjacent) as
explained below.
If xn(n-1)⋅⋅⋅5-1=c2, a healthy path (c2, c1, c0) in Yn(n-1)⋅⋅⋅5-1
together with all healthy paths that we have established in
A0, A1, …, An(n-1)⋅⋅⋅5-2 form a ring of odd length in Sn. This is
a contradiction because Sn is a bipartite graph in which
every ring has even length. Similarly, there is a contradiction for xn(n-1)⋅⋅⋅5-1=c4. Now that xn(n-1)⋅⋅⋅5-1 and y n(n-1)⋅⋅⋅5-1 are adjacent, there is a healthy hamiltonian xn(n-1)⋅⋅⋅5-1
-yn (n-1)⋅⋅⋅5-1 path in An(n-1)⋅⋅⋅5-1 if An(n-1)⋅⋅⋅5-1 is healthy. If A
n(n-1)⋅⋅⋅5-1 is faulty, there is a healthy xn(n-1)⋅⋅⋅5-1-yn(n-1)⋅⋅⋅5-1 path of
maximal length 4!-3 in An(n-1)⋅⋅⋅5-1 according to Lemma 4.
When Xn(n-1)⋅⋅⋅5-1≠Yn(n-1)⋅⋅⋅5-1, a healthy hamiltonian X
n(n-1)⋅⋅⋅5-1-Yn(n-1)⋅⋅⋅5-1 path for the
3 4
K
formed by An(n-1)⋅⋅⋅5-1 isfirst established, similar to the situation of A0. Without loss
of generality, we assume (Xn(n-1)⋅⋅⋅5-1, D, E, Yn(n-1)⋅⋅⋅5-1) is the
healthy hamiltonian Xn(n-1)⋅⋅⋅5-1-Yn(n-1)⋅⋅⋅5-1 path. There are
three pairs of distinct healthy vertices, denoted by {x n(n-1)⋅⋅⋅5-1, a}, {b, c}, and {d, e}, in Xn(n-1)⋅⋅⋅5-1, D, and E,
respectively, so that every two consecutive vertices in
{xn(n-1)⋅⋅⋅5-1, a, b, c, d, e, f} are adjacent, where f is a healthy
vertex of Yn(n-1)⋅⋅⋅5-1. Since yn(n-1)⋅⋅⋅5-1=c0, f cannot be c0 or c3
according to Lemmas 5 and 6. Moreover, f ∈{c2, c4} will
result in a ring of odd length in Sn, which is a contradiction.
Hence, f is adjacent to yn(n-1)⋅⋅⋅5-1 (i.e., f=c1 or c5). It is easy
to establish a healthy hamiltonian xn(n-1) ⋅⋅⋅5-1-yn(n-1)⋅⋅⋅5-1 path
in An(n-1)⋅⋅⋅5-1.
Clearly, all the healthy paths that we have established in A0, A1, …, An(n-1)⋅⋅⋅5-1 together with edges (y0, x1), (y1, x2), …, (yn(n-1)⋅⋅⋅5-1, x0) constitute a healthy ring of length
n!-2|Fv|. Since Sn is bipartite with two partite sets of equal size,
the ring is the longest in the worst case. Q.E.D.
Theorem 1. Suppose n≥4 and |Fv|≤n-3. Sn can embed a
healthy ring of maximal length n!-2|Fv|.
Proof. We show that Sn contains a healthy ring of
maximal length n!-2|Fv| according to different values of n.
When n=4, |Fv|≤1. By the aid of Lemma 4, there is a
healthy ring of maximal length 4!-2|Fv| in S4.
When n=5, |Fv|≤2. Without loss of generality, we
assume |Fv|=2. According to Lemma 2, the two vertex
faults of S5 will be distributed to two different 4-vertices
after an a1-partition is executed, where 2≤a1≤5. Then an R4
is established so that the two faulty 4-vertices are not adjacent in the R4. Clearly the R4 satisfies properties (P1)
and (P3). Besides, it is not difficult to check that the R4
also satisfies property (P2). By Lemma 7, a healthy ring of maximal length 5!-2|Fv| can be generated from the R4.
When n≥6, an R4 with properties (P1), (P2), and (P3)
is first obtained by Lemma 3. Then, a healthy ring of maximal length n!-2|Fv| can be generated by Lemma 7.
Q.E.D.
In this paper, we have shown that a healthy ring of length
n!-2|Fv| can be embedded onto an n-dimensional star graph
with |Fv|≤n-3 vertex faults. This improves Tseng et al.’s
result by lengthening the embedded ring from n!-4|Fv| to n!-2|Fv|. Since the star graph is regular of degree n-1 and is
bipartite with two partite sets of equal size, our result achieves optimality in the worst case.
In [32], Tseng et al. also showed that an n-dimensional star graph with |Fv| vertex faults and |Fe| edge
faults can embed a healthy ring of length at least n!-4|Fv| if
|Fv|+|Fe|≤n-3. With our result, the embedded ring can be
lengthened from n!-4|Fv| to n!-2|Fv| as well.
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